Find $\dfrac{d}{dx}\left[ 2x\sqrt{x}\sin(x) \right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\cos(x)}{\sqrt{x}}$ (Choice B) B $3\sqrt{x}\sin(x)+2x\sqrt{x}\cos(x)$ (Choice C) C $\dfrac{\sin(x)+2x\cos(x)}{\sqrt{x}}$ (Choice D) D $\dfrac{2\sin(x)+x\cos(x)}{\sqrt{\sin(x)}}$
$2x\sqrt{x}\sin(x)$ is a product of three functions. Let... $u(x)=2x$ $v(x)=\sqrt{x}$ $w(x)=\sin(x)$... then $2x\sqrt{x}\sin(x)=u(x)\cdot v(x) \cdot w(x)$. To find $\dfrac{d}{dx}\left[ 2x\sqrt{x}\sin(x) \right]$, we will need to use the product rule twice! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u(x)\cdot v(x) \cdot w(x)\right] \\\\ &=u'(x)\cdot \Bigl(v(x)\cdot w(x) \Bigr)+u(x)\cdot\dfrac{d}{dx}\left[v(x)\cdot w(x)\right]&\gray{\text{Product rule}} \\\\ &=u'(x)\cdot \Bigl(v(x)\cdot w(x) \Bigr)+u(x)\cdot\Bigl(v'(x)\cdot w(x)+v(x)w'(x)\Bigr)&\gray{\text{Product rule}} \\\\ &= u'(x)\cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=2$ $v'(x)=\dfrac{1}{2\sqrt{x}}$ $w'(x)=\cos(x)$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}u'(x)\cdot v(x) \cdot w(x) + u(x) \cdot v'(x) \cdot w(x) + u(x) \cdot v(x) \cdot w'(x) \\\\ &=2 \cdot \sqrt{x} \cdot \sin(x) + 2x \cdot \dfrac{1}{2\sqrt{x}} \cdot \sin(x) + 2x \cdot \sqrt{x} \cdot \cos(x) \\\\ &=3\sqrt{x}\sin(x)+2x\sqrt{x}\cos(x) \end{aligned}$ In conclusion: $\dfrac{d}{dx}\left[ 2x\sqrt{x}\sin(x) \right]=3\sqrt{x}\sin(x)+2x\sqrt{x}\cos(x)$